\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)

a, Theo PT: \(n_{CH_3COOH}=2n_{Fe}=0,2\left(mol\right)\Rightarrow m_{CH_3COOH}=0,2.60=12\left(g\right)\)

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{12}{10\%}=120\left(g\right)\)

\(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(n_{\left(CH_3COO\right)_2Fe}=n_{Fe}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 5,6 + 120 - 0,1.2 = 125,4 (g)

\(\Rightarrow C\%_{\left(CH_3COO\right)_2Fe}=\dfrac{0,1.174}{125,4}.100\%\approx13,88\%\)

\(m_{CH_3COOH}=24\%.150=36\left(g\right)\\ \rightarrow n_{CH_3COOH}=\dfrac{36}{60}=0,6\left(mol\right)\)

PTHH: 2CH₃COOH + Na₂CO₃ ---> 2CH₃COONa + CO₂ + H₂O

                0,6                  0,3                  0,6                  0,3 

=> V_CO2 = 0,3.22,4 = 6,72 (l)

\(m_{Na_2CO_3}=0,3.31,8\left(g\right)\)

=> \(m_{ddNa_2CO_3}=\dfrac{31,8}{21,2\%}=150\left(g\right)\)

m_CO2 = 0,3.44 = 13,2 (g)

\(m_{dd}=150+150-13,2=286,8\left(g\right)\)

\(m_{CH_3COONa}=0,3.82=24,6\left(g\right)\\ \rightarrow C\%_{CH_3COONa}=\dfrac{24,6}{286,8}=8,58\%\)

a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PT: \(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, \(n_{\left(CH_3COO\right)_2Zn}=n_{Zn}=0,1\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Zn}=0,1.183=18,3\left(g\right)\)

c, \(n_{CH_3COOH}=2n_{Zn}=0,2\left(mol\right)\)

\(\Rightarrow m_{CH_3COOH}=0,2.60=12\left(g\right)\)

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{12}{60\%}=20\left(g\right)\)

a) \(n_{HCl}=0,1.2=0,2\left(mol\right)\)

PTHH: CuO + 2HCl --> CuCl₂ + H₂O

______0,1<---0,2------->0,1

=> a = 0,1.80 = 8(g)

b) \(C_{M\left(CuCl_2\right)}=\dfrac{0,1}{0,1}=1M\)

c) 

PTHH: CuCl₂ + 2NaOH --> Cu(OH)₂ + 2NaCl

______0,1------------------------>0,1

=> mCu(OH)₂ = 0,1.98 = 9,8(g)

1)

a, \(n_{Al}=\dfrac{15,3}{102}=0,15\left(mol\right)\)

PTHH: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

Mol:     0,15         0,9        0,3      

\(m_{ddHCl}=\dfrac{0,9.36,5.100}{20}=164,25\left(g\right)\)

b, m_{dd sau pứ} = 15,3 + 164,25 = 179,55 (g)

c, \(C\%_{ddAlCl_3}=\dfrac{0,3.133,5.100\%}{179,55}=22,31\%\)

2)

a, \(m_{HCl}=54,75.20\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: Al₂O₃  + 6HCl → 2AlCl₃ + 3H₂O

Mol:      0,05        0,3          0,1

\(m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)

b, m_{dd sau pứ} = 5,1 + 54,75 = 59,85 (g) 

\(C\%_{ddAlCl_3}=\dfrac{0,1.133,5.100\%}{59,85}=22,31\%\)